Calculus 3: Work Homework: Electric Dipole Field

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SUMMARY

The discussion focuses on calculating the work required to move a charge in the electric field generated by an electric dipole with a dipole moment of p = 4 × 10-5 C-m. The electric field is defined by the formula F(x,y,z) = kp/r5 <3xz, 3yz, 2z2 - x2 - y2 >, where k = 8.99 × 109 N-m2/C2. The work done against the force F to move a charge q = 0.01 C from the point (1,−5, 0) to (3, 4, 4) can be calculated directly from the force without needing to find a potential function, as the path is not specified in the problem.

PREREQUISITES
  • Understanding of electric dipole moment and its implications in electric fields.
  • Familiarity with vector calculus, particularly gradient and parametric equations.
  • Knowledge of work-energy principles in the context of electric fields.
  • Proficiency in evaluating line integrals, specifically ∫ F(c(t)) * c'(t) dt.
NEXT STEPS
  • Study the derivation and implications of electric dipole fields in three dimensions.
  • Learn how to calculate work done by a force field using line integrals.
  • Explore the concept of potential functions in vector fields and their applications.
  • Investigate different path parametrizations and their effects on work calculations in electric fields.
USEFUL FOR

Students and professionals in physics, particularly those studying electromagnetism, as well as anyone involved in solving problems related to electric fields and work calculations in vector calculus.

tarantino5
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Homework Statement



An electric dipole with dipole moment p = 4 × 10−5 C-m sets up
an electric field (in Newtons per coulomb)

F(x,y,z) = kp/r^5 <3xz, 3yz, 2z^{2} - x^{2} - y^{2}>

where r = (x^{2} + y^{2} + z^{2})^{1/2} with distance in meters and k = 8.99 ×
109 N-m^{2}/C^{2}. Calculate the work against F required to move a particle
of charge q = 0.01 C from (1,−5, 0) to (3, 4, 4). Note: The force on
q is qF Newtons.

Homework Equations



∫^{b}_{a} F(c(t)) * c'(t) dt

The Attempt at a Solution



I know that I need to set ∇f = F and find a potential function. I'm not sure what the parametric equation should be.
 
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tarantino5 said:
I know that I need to set ∇f = F and find a potential function. I'm not sure what the parametric equation should be.

Perhaps you don't need to calculate the potential at all. Seems like you get work directly from the force, according to your formula.

The path parametrization depends on the path of course. It seems a path is not specified in the problem. Why is it not needed to solve the problem?
 

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